Question 375927
Since they both equal y:
{{{log(2,(x+1)) = 5 - log(2,(x-3))}}}
{{{log(2,(x+1)) = log(2,32) - log(2,(x-3))}}}
{{{log(2,(x+1)) = log(2,32/(x-3))}}}
x+1 = 32/(x-3)
{{{x^2 - 2x - 3 = 32}}}
{{{x^2 - 2x - 35 = 0}}}
(x-7)*(x+5) = 0
x = -5  Discard, since it gives the log of a negative number
x = 7
-------
{{{y = log(2,(7+1)) = log(2,8)}}}
y = 3
-----
(x,y) = (7,3)